BIOL 1090 Lecture Notes - Lecture 4: Dihybrid Cross, Punnett Square, Mendelian Inheritance

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Recall from last lecture we focused on generating gametes from 2 genes, each with 2 alleles using. Step 2: the f1 heterozygotes produce four kinds of gametes in equal proportions gw gw gw and gw. Re(cid:272)all (cid:1007): me(cid:374)del"s pri(cid:374)(cid:272)iple of i(cid:374)depe(cid:374)de(cid:374)t assort(cid:373)e(cid:374)t: alleles on different pairs of chromosomes assort independently from one another, because of this rule, we can think of each trait as independent of the other. Therefore, we can just look at one gene at a time: homologous chromosomes disjoin and move to opposite poles of the cell anaphase i. Review: predicting the outcomes of a monohybrid cross with a punnet square: for gene 1 with 2 alleles they are: 2^1 = 2 possibble haploid genotypes (gametes: combining 2 different gametes from each of two parents: 2x2 = Recall for 1 gene with 2 alleles there are 2 possibple haploid genotype. If one parent was homozygous, he/she would generate only 1 kind of gamete.

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